# INSTANTANEOUS ACCELERATION FORMULA AND EXAMPLE

## Instantaneous Acceleration Formula

Instantaneous acceleration is defined as the acceleration at any moment of time and instantaneous acceleration formula is

$${ a }_{ t }=lim_{ \Delta t\rightarrow 0 }\frac { { v}_{ { t }+\Delta t }-{ v }_{ t } }{ \Delta t } =\frac { dv }{ dt }$$.

As seen from formula, instantaneous acceleration will be the limit for ∆t goes to zero for v measured at t+∆t minus vt divided by ∆t. This is the first derivative of velocity vs. time and which is also second derivative of position vs. time.

We can look at the Fig. 1 and find where is the acceleration is zero, positive and negative. Since the plot is position versus time and acceleration is the second derivative of position versus time, we need to see how the slope is changing .

## Instantaneous Acceleration Example

A position of an object at any moment in time is defined by the equation

$$x=8-6t+{ t }^{ 2 }\quad m.$$

What is the velocity and acceleration at any moment in time?

The velocity at any time t is

$$v=\frac { dx }{ dt } =-6+2t\quad \frac { m }{ s } .$$

The instantaneous acceleration is

$$v=\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } =2\quad \frac { m }{ { s }^{ 2 } } .$$

The acceleration is constant in time and it’s not changing but the velocity is changing.

What is the position, velocity and acceleration at time t=0 ?

$$x=8-6t+{ t }^{ 2 }=8-6(0)+{ (0) }^{ 2 }=8{ m }$$,

$$v=-6+2t=-6+2(0)=-6 { m }/{ s }$$,

$$a=2\quad { m }/{ { s }^{ 2 } }$$.

At what time, x = 0? When we solve position equation for x=0,

$$x=8-6t+{ t }^{ 2 }=(t-4)(t-2)=0$$

we find that at time t=2 s and t=4 s, the position x is 0.

When the velocity is is 0? When we solve velocity equation for v=0,

$$v=-6+2t=0\rightarrow t=6/2=3$$

we find that at time t=3 s the velocity is 0.

The plot x as a function time for this example is a parabola and given below.

When we look at the plot in Fig 2, we see that object starts out at t=0 with negative velocity but acceleration is in positive direction. For this reason, velocity is slowly changing and that comes a time where velocity is 0 (t=3). Velocity changes from negative to positive because of the positive acceleration. At time t=4, the velocity is v=-6+2×4=2 m/s and it’s positive.